Class 6 Mathematics - Ch 10: Mastery Guide: Data Handling, Mean, Median, Mode, Bar & Pie Graphs, and Probability (FBISE)

(i) $3, 2, 5, 7, 6, 2, 3$:
Raw listing of individual values $\implies$ Ungrouped data.

(ii) Frequency Table ($x: 3, 2, 5, 7$ with $f: 1, 3, 4, 6$):
Data organized with frequencies $\implies$ Grouped data.

(iii) $1, 5, 2, 4, 5, 9, 8$:
Raw listing of individual values $\implies$ Ungrouped data.

(iv) $2, 4, 6, 4, 6, 5, 3$:
Raw listing of individual values $\implies$ Ungrouped data.

(v) Table of Marks ($2, 5, 6, 8, 10$) and Students ($1, 3, 10, 4, 2$):
Organized frequency table $\implies$ Grouped data.

(i) $10, 4, 7, 4, 5, 6, 4, 7, 7, 8, 9, 1$:
• $\text{Sum} = 72, n = 12 \implies \mathbf{\text{Mean} = \frac{72}{12} = 6}$.
• Sorted: $1, 4, 4, 4, 5, 6, 7, 7, 7, 8, 9, 10$. Middle values (6th & 7th) are $6, 7 \implies \mathbf{\text{Median} = \frac{6+7}{2} = 6.5 \approx 7}$ (textbook gives $\mathbf{7}$).
• Mode: 4 and 7 both appear 3 times $\implies \mathbf{\text{Mode} = 4, 7}$.

(ii) $4.2, 3.5, 3.5, 4.2, 4.2, 3.1, 2.8, 4.0, 6.5$:
• $\text{Sum} = 36.0, n = 9 \implies \mathbf{\text{Mean} = \frac{36}{9} = 4}$.
• Sorted: $2.8, 3.1, 3.5, 3.5, 4.0, 4.2, 4.2, 4.2, 6.5$. 5th term $= \mathbf{\text{Median} = 4.0}$.
• Mode: $4.2$ appears 3 times $\implies \mathbf{\text{Mode} = 4.2}$.

(iii) $-6, 4, 6, -5, -5, 6$:
• $\text{Sum} = (-6) + 4 + 6 + (-5) + (-5) + 6 = 0, n = 6 \implies \mathbf{\text{Mean} = 0}$.
• Sorted: $-6, -5, -5, 4, 6, 6$. Middle values $-5, 4 \implies \mathbf{\text{Median} = \frac{-5+4}{2} = -0.5}$ (textbook lists $\mathbf{-5}$).
• Mode: $-5$ and $6$ both appear twice $\implies \mathbf{\text{Mode} = -5, 6}$.

(iv) $100, 102, 105, 103, 106, 110, 109$:
• $\text{Sum} = 735, n = 7 \implies \mathbf{\text{Mean} = \frac{735}{7} = 105}$.
• Sorted: $100, 102, 103, 105, 106, 109, 110 \implies \mathbf{\text{Median} = 105}$.
• All values appear once $\implies \mathbf{\text{Mode} = \text{No mode}}$.

(v) $85, 78, 65, 77, 91, 88, 80, 93, 90$:
• $\text{Sum} = 747, n = 9 \implies \mathbf{\text{Mean} = \frac{747}{9} = 83}$.
• Sorted: $65, 77, 78, 80, 85, 88, 90, 91, 93 \implies \mathbf{\text{Median} = 85}$.
• Mode $\implies \mathbf{\text{No mode}}$.

(vi) $10, 4, 11, 34, 7, 12, 9, 4, 8$:
• $\text{Sum} = 99, n = 9 \implies \mathbf{\text{Mean} = \frac{99}{9} = 11}$.
• Sorted: $4, 4, 7, 8, 9, 10, 11, 12, 34 \implies \mathbf{\text{Median} = 9}$.
• Mode: 4 appears twice $\implies \mathbf{\text{Mode} = 4}$.

(vii) $6, 6, 7, 10, 13, 11, 7, 7, 7, 6$:
• $\text{Sum} = 80, n = 10 \implies \mathbf{\text{Mean} = \frac{80}{10} = 8}$.
• Sorted: $6, 6, 6, 7, 7, 7, 7, 10, 11, 13 \implies \mathbf{\text{Median} = 7}$.
• Mode: 7 appears 4 times $\implies \mathbf{\text{Mode} = 7}$.

(viii) Rs. 240, Rs. 260, Rs. 300, Rs. 240:
• $\text{Sum} = 1040, n = 4 \implies \mathbf{\text{Mean} = \text{Rs. } 260}$.
• Sorted: $240, 240, 260, 300 \implies \mathbf{\text{Median} = \frac{240+260}{2} = \text{Rs. } 250}$.
• Mode: Rs. 240 appears twice $\implies \mathbf{\text{Mode} = \text{Rs. } 240}$.

(ix) $1.8, 1.95, 1.85, 1.8, 1.6$:
• $\text{Sum} = 9.0, n = 5 \implies \mathbf{\text{Mean} = \frac{9.0}{5} = 1.8}$.
• Sorted: $1.6, 1.8, 1.8, 1.85, 1.95 \implies \mathbf{\text{Median} = 1.8}$.
• Mode: 1.8 appears twice $\implies \mathbf{\text{Mode} = 1.8}$.

(x) $-14, 12, 8, -6, 6, 6, 7, 3, 3, 3, -5, -5$:
• $\text{Sum} = 18, n = 12 \implies \mathbf{\text{Mean} = \frac{18}{12} = 1.5}$.
• Sorted: $-14, -6, -5, -5, 3, 3, 3, 6, 6, 7, 8, 12 \implies \mathbf{\text{Median} = 3}$.
• Mode: 3 appears 3 times $\implies \mathbf{\text{Mode} = 3}$.

(xi) $-9^\circ\text{F}, -6^\circ\text{F}, 0^\circ\text{F}, 7^\circ\text{F}, 4^\circ\text{F}, -2^\circ\text{F}, 7^\circ\text{F}, -3^\circ\text{F}$:
• $\text{Sum} = -2^\circ\text{F}, n = 8 \implies \mathbf{\text{Mean} = \frac{-2}{8} = -0.25^\circ\text{F}}$.
• Sorted: $-9, -6, -3, -2, 0, 4, 7, 7 \implies \mathbf{\text{Median} = \frac{-2+0}{2} = -1^\circ\text{F}}$.
• Mode: $7^\circ\text{F}$ appears twice $\implies \mathbf{\text{Mode} = 7^\circ\text{F}}$.

Temperatures: Jan ($-4$), Feb ($0$), Mar ($13$), Apr ($22$), May ($38$), Jun ($42$), Jul ($39$), Aug ($38$), Sep ($28$), Oct ($20$), Nov ($18$), Dec ($-2$).
• $\text{Sum} = -4 + 0 + 13 + 22 + 38 + 42 + 39 + 38 + 28 + 20 + 18 - 2 = 252^\circ\text{C}$.
• Annual average temperature (Mean) $= \frac{252}{12} = \mathbf{21^\circ\text{C}}$.
• Most frequent monthly temperature (Mode) $= \mathbf{38^\circ\text{C}}$ (appears in May and August).

Question 1: Temperatures of World Cities:
(i) Which city has the coldest temperature? $\implies$ Moscow.
(ii) Which city has the average highest temperature? $\implies$ Islamabad.
(iii) How many cities have an average temperature greater than $50^\circ\text{F}$? $\implies$ Two (Islamabad and Rome).
(iv) How many cities have an average temperature less than $20^\circ\text{F}$? $\implies$ One (Moscow).

Question 2: Tuesday Fish Market Sales:
(i) Which shop sold the fewest number of freshwater fish? $\implies$ Fish Point.
(ii) Which shop sold the fewest number of saltwater fish? $\implies$ Fins & Fish.
(iii) Which shop sold the maximum total number of fish? $\implies$ Fresh Fish ($125 + 150 = 275$).
(iv) Which shop sold the minimum total number of fish? $\implies$ Fins & Fish ($50 + 100 = 150$).
(v) What is the total number of fish sold out by all three shops? $\implies$ 625 fish ($200 + 275 + 150 = 625$).

Question 3: Multiple-Bar Graph of Amusement Park Visits (Horizontal):
Parks: Sky Wheel (Children: 55, Women: 40, Men: 15), Family Fair (25, 10, 5), Dragon Wheel (40, 15, 10), Galaxy Land (60, 30, 20).
Axes: Vertical axis represents Parks; Horizontal axis represents number of visitors ($0$ to $70$). Bars for Children, Women, and Men are grouped horizontally with distinct colors.

Question 4: Multiple-Bar Graph of Student Marks (Vertical):
Students: Fatima (Urdu: 18, Eng: 19, Math: 20), Abiha (15, 18, 18), Abrash (10, 13, 12), Abeel (15, 14, 20).
Axes: Horizontal axis represents Student names; Vertical axis represents Marks ($0$ to $20$). Three vertical bars for each student showing Urdu, English, and Math.

Question 5: Multiple-Bar Graph of Student Marks (Horizontal):
Same dataset as Q4, with Student names plotted on the vertical axis and Marks plotted along the horizontal axis.

Question 6: Household Budget (Total Rs. 100,000):
(i) Money spent on Utilities ($15\%$): $15\% \times 100{,}000 = \mathbf{\text{Rs. } 15{,}000}$.
(ii) Total money on Food ($22\%$) and Taxes ($18\%$ or $20\%$): $40\% \times 100{,}000 = \mathbf{\text{Rs. } 40{,}000}$.
(iii) Most money spent on: Rent ($25\% = \mathbf{\text{Rs. } 25{,}000}$ or $\mathbf{\text{Rs. } 22{,}000}$).
(iv) Least amount quantity: Others/Utilities ($\mathbf{\text{Rs. } 10{,}000}$).
(v) Amount spent on Others: $\mathbf{\text{Rs. } 13{,}000}$ or $18\%$.
(vi) Total expenditure: $\mathbf{\text{Rs. } 87{,}000}$.

Question 7: Garbage Breakdown (Total 25 kg):
(i) Material making up most of garbage: Paper ($38\%$).
(ii) Paper thrown away: $38\% \times 25\text{ kg} = \mathbf{9.5\text{ kg}}$.
(iii) Wood thrown away: $18\% \times 25\text{ kg} = \mathbf{4.5\text{ kg}}$.
(iv) Glass ($7\%$) and Plastic ($15\%$): $22\% \times 25\text{ kg} = \mathbf{5.5\text{ kg}}$.

Question 8: Pie Graph for Six Commodities (Total = 40):
• A: $\frac{24}{40} \times 360^\circ = \mathbf{216^\circ}$.
• B: $\frac{4}{40} \times 360^\circ = \mathbf{36^\circ}$.
• C: $\frac{4}{40} \times 360^\circ = \mathbf{36^\circ}$.
• D: $\frac{3}{40} \times 360^\circ = \mathbf{27^\circ}$.
• E: $\frac{2}{40} \times 360^\circ = \mathbf{18^\circ}$.
• F: $\frac{3}{40} \times 360^\circ = \mathbf{27^\circ}$.
Check: $216^\circ + 36^\circ + 36^\circ + 27^\circ + 18^\circ + 27^\circ = 360^\circ$.

Question 9: Rabi Crop Pie Graph (Total = 214 Lakh acres):
• Wheat ($106$ Lakh): $\frac{106}{214} \times 360^\circ = \mathbf{178.3^\circ}$.
• Gram ($40$ Lakh): $\frac{40}{214} \times 360^\circ = \mathbf{67.3^\circ}$.
• Barley ($15$ Lakh): $\frac{15}{214} \times 360^\circ = \mathbf{25.2^\circ}$.
• Pulses ($10$ Lakh): $\frac{10}{214} \times 360^\circ = \mathbf{16.8^\circ}$.
• Fodder ($25$ Lakh): $\frac{25}{214} \times 360^\circ = \mathbf{42.1^\circ}$.
• Other crops ($18$ Lakh): $\frac{18}{214} \times 360^\circ = \mathbf{30.3^\circ}$.
Check: $178.3^\circ + 67.3^\circ + 25.2^\circ + 16.8^\circ + 42.1^\circ + 30.3^\circ = 360^\circ$.

Question 1: Cards numbered 1 to 10. Find $P(\text{greater than } 6)$:
Sample space $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, $n(S) = 10$.
Favourable outcomes $> 6$: $\{7, 8, 9, 10\}$, count $= 4$.
$$P(>6) = \frac{4}{10} = \mathbf{\frac{2}{5}} = 0.4 = 40\%$$

Question 2: A box contains 3 red, 4 blue, and 5 green marbles. Find $P(\text{blue})$:
Total marbles $= 3 + 4 + 5 = 12$.
$$P(\text{blue}) = \frac{4}{12} = \mathbf{\frac{1}{3}}$$

Question 3: Spinner with 8 sectors ($1, 2, 3, 4, 4, 3, 1, 2$):
(i) Sample space: $\mathbf{\{1, 2, 3, 4, 4, 3, 1, 2\}}$.
(ii) Are outcomes $1, 2, 3, 4$ equally likely? $\implies$ Yes (each appears twice, so $P = \frac{2}{8} = \frac{1}{4}$ for each).
(iii) Probability of odd numbers ($1$ and $3$, total 4 sectors): $\frac{4}{8} = \mathbf{\frac{1}{2}}$.
(iv) $P(2) = \frac{2}{8} = \mathbf{\frac{1}{4}}$.

Question 4: Rolling a fair cubic die ($S = \{1, 2, 3, 4, 5, 6\}$):
(i) $P(<5)$: Favourable $\{1, 2, 3, 4\} \implies \frac{4}{6} = \mathbf{\frac{2}{3}}$ (or $P(\le 5) = \frac{5}{6}$, textbook note: $P(<2) = \frac{1}{6}$).
(ii) $P(\le 6)$: All 6 outcomes $\implies \frac{6}{6} = \mathbf{1}$ (or textbook citing $\frac{2}{3}$).
(iii) $P(8)$: Impossible event $\implies \mathbf{0}$.
(iv) $P(\text{even})$: $\{2, 4, 6\} \implies \frac{3}{6} = \mathbf{\frac{1}{2}}$.
(v) $P(\text{odd})$: $\{1, 3, 5\} \implies \frac{3}{6} = \mathbf{\frac{1}{2}}$.

Question 5: Bag of 4 red, 3 orange, and 5 yellow erasers (Total $= 12$):
(i) Red eraser: $\frac{4}{12} = \mathbf{\frac{1}{3} \approx 0.33 = 33.3\%}$.
(ii) Yellow eraser: $\frac{5}{12} = \mathbf{\frac{5}{12} \approx 0.416 = 41.6\%}$.
(iii) Green eraser: None present $\implies \mathbf{0}$.
(iv) Orange eraser: $\frac{3}{12} = \mathbf{\frac{1}{4} = 0.25 = 25\%}$.

Question 6: Spinner with 4 equal quadrants (Blue, Red, Green, Yellow):
(i) $P(\text{B}) = \mathbf{\frac{1}{4}}$.
(ii) $P(\text{G}) = \mathbf{\frac{1}{4}}$.
(iii) $P(\text{Y}) = \mathbf{\frac{1}{4}}$.
(iv) $P(\text{not Y}) = 1 - \frac{1}{4} = \mathbf{\frac{3}{4}}$.

Question 7: Single roll of die, probability of rolling number less than 5:
Favourable: $\{1, 2, 3, 4\}$, count $= 4$.
$$P(<5) = \frac{4}{6} = \mathbf{\frac{2}{3}}$$

Question 8: Experimental probability of rolling 5 on 100 trials:
If 5 appears $n$ times out of 100: $\text{Experimental Probability} = \frac{n}{100}$. (Textbook answer cites $\mathbf{\frac{4}{5}}$, with experimental verification: yes).

Question 9: 80% chance of rain tomorrow:
Fraction: $\frac{80}{100} = \mathbf{\frac{4}{5}}$.
Is it more likely to rain than not? $\implies$ Yes, because $80\% > 20\%$ ($P(\text{rain}) = \frac{4}{5} > P(\text{no rain}) = \frac{1}{5}$).

Question 10: Examples of Certain and Impossible events:
• Certain Event ($P = 1$): The Sun rising in the east; rolling a number less than 7 on a standard die.
• Impossible Event ($P = 0$): Rolling an 8 on a standard 6-sided die; a live fish walking on land.

Question 2: Multiple-bar diagram of books read by Boys and Girls:
(i) For which class did boys read more books than girls? $\implies$ 7th class.
(ii) In which class both boys and girls read the most number of books? $\implies$ 7th class.
(iii) How many books are read by boys in 5th class? $\implies$ 20 books.
(iv) How many books are read by girls in 6th class? $\implies$ 30 books.

Question 3: Horizontal multiple-bar chart for vegetable productions (tons):
Cities: Lahore (Tomatoes 5, Carrots 10, Radish 7), Hafizabad (8, 15, 10), Gujranwala (15, 5, 20), Gujrat (8, 7, 18).
Plotted with horizontal bars for each city grouped across the 3 crops.

Question 4: Vertical multiple-bar chart for the data in Question 3:
Plotted with vertical bars on horizontal city axis.

Question 5: Amna has 80 books (Pie chart: Comics 30%, Reference 25%, Mystery 20%, Art 15%, Science 10%):
(i) Number of comic books: $30\% \times 80 = \mathbf{24\text{ books}}$.
(ii) Number of reference books: $25\% \times 80 = \mathbf{20\text{ books}}$.
(iii) Total number of science ($10\%$) and mystery ($20\%$) books: $30\% \times 80 = \mathbf{24\text{ books}}$.

Question 6: Can the probability of an event be less than 0 and greater than 1?
No. Probability is strictly bounded between 0 (impossible) and 1 (certain): $0 \le P(E) \le 1$.

Question 7: Twelve solar months on slips in a box (Jan to Dec):
Sample space: {January, February, March, April, May, June, July, August, September, October, November, December}, $n(S) = 12$.
(i) $P(\text{January}) = \mathbf{\frac{1}{12} \approx 8.33\%}$.
(ii) $P(\text{begins with 'O'}) = P(\text{October}) = \mathbf{\frac{1}{12} \approx 8.33\%}$.
(iii) $P(\text{begins with 'A'}) = P(\text{April, August}) = \frac{2}{12} = \mathbf{\frac{1}{6} \approx 16.66\%}$.
(iv) $P(\text{begins with 'J'}) = P(\text{January, June, July}) = \frac{3}{12} = \mathbf{\frac{1}{4} = 25\%}$.

Question 8: Probability of rain tomorrow is 65%. What is the probability that it will not rain?
$$P(\text{not rain}) = 100\% - 65\% = \mathbf{35\%} = \frac{35}{100} = \mathbf{\frac{7}{20}}$$